Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

The initial velocity of a body moving along a straight line is $7 \text{ m/s}$ . It has a uniform acceleration of $4 \text{ m/s}^2$ . The distance covered by the body in the 5th second of its motion is:

25 m

35 m

50 m

85 m

Step-by-Step Solution

Identify Formula: The distance covered in the $n$ -th second of uniformly accelerated motion is given by the formula: $S_n = u + \frac{a}{2}(2n - 1)$ , where $u$ is initial velocity, $a$ is acceleration, and $n$ is the specific second .
Alternative Method: Calculate the displacement at $t=5$ and $t=4$ using $S = ut + \frac{1}{2}at^2$ and find the difference. $S(5) = 7(5) + \frac{1}{2}(4)(5^2) = 35 + 2(25) = 85 \text{ m}$ . $S(4) = 7(4) + \frac{1}{2}(4)(4^2) = 28 + 2(16) = 60 \text{ m}$ .

Distance in 5th second $= S(5) - S(4) = 85 - 60 = 25 \text{ m}$ .

Direct Calculation: Using the $n$ -th second formula: $S_5 = 7 + \frac{4}{2}(2 \times 5 - 1)$ $S_5 = 7 + 2(10 - 1) = 7 + 2(9) = 7 + 18 = 25 \text{ m}$ .

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